Преобразование выражения, содержащего производную

[never-sleep]

Подскажите, плиз, как тут быть с такой задачей?

Преобразовать $z''_{xy}$, сделав замену переменных

$x=u\quad\quad y=w\quad\quad z=uwv^2$,

где $u=u(v,w)$

Верно ли начал решать?

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial w}\dfrac{\partial w}{\partial y}+\dfrac{\partial z}{\partial v}\cdot \dfrac{\partial v}{\partial y}$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial w}\cdot 1+\dfrac{\partial z}{\partial v}\cdot 1$

$\dfrac{\partial z}{\partial w}=\dfrac{\partial uwv^2}{\partial w}=uwv^2+wv^2\dfrac{\partial u}{\partial w}$

$\dfrac{\partial z}{\partial u}=\dfrac{\partial uwv^2}{\partial v}=2vwu+wv^2\dfrac{\partial u}{\partial v}$

$\Phi\left(u,v,w,\dfrac{\partial u}{\partial w},\dfrac{\partial u}{\partial v}\right)=\dfrac{\partial z}{\partial y}=uwv^2+wv^2\dfrac{\partial u}{\partial w}+2vwu+wv^2\dfrac{\partial u}{\partial v}$

$z''_{xy}=\dfrac{\partial}{\partial x} \left(uwv^2+wv^2\dfrac{\partial u}{\partial w}+2vwu+wv^2\dfrac{\partial u}{\partial v} \right)$

$z''_{xy}=\dfrac{\partial \Phi}{\partial w}\cdot \dfrac{\partial w}{\partial x}+\dfrac{\partial \Phi}{\partial u}\cdot \dfrac{\partial u}{\partial x}$